Download Chapter 5 Forces and Motion II

Chapter 5
Forces and Motion II
5.1
5.1.1
The Important Stuff
Friction Forces
Forces which are known collectively as “friction forces” are all around us in daily life. In
elementary physics we discuss the friction force as it occurs between two objects whose
surfaces are in contact and which slide against one another.
If in such a situation, a body is not moving while an applied force F acts on it, then
static friction forces are opposing the applied force, resulting in zero net force. Empirically,
one finds that this force can have a maximum value given by:
fsmax = µs N
(5.1)
where µs is the coefficient of static friction for the two surfaces and N is the normal
(perpendicular) force between the two surfaces.
If one object is in motion relative to the other one (i.e. it is sliding on the surface) then
there is a force of kinetic friction between the two objects. The direction of this force is
such as to oppose the sliding motion and its magnitude is given by
fk = µk N
(5.2)
where again N is the normal force between the two objects and µk is the coefficient of
kinetic friction for the two surfaces.
5.1.2
Uniform Circular Motion Revisited
Recall the result given in Chapter 3: When an object is in uniform circular motion, moving
in a circle of radius r with speed v, the acceleration is directed toward the center of the circle
and has magnitude
acent =
103
v2
.
r
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CHAPTER 5. FORCES AND MOTION II
Therefore, by Newton’s Second Law of Motion, the net force on this object must also be
directed toward the center of the circle and have magnitude
Fcent =
mv 2
.
r
(5.3)
Such a force is called a centripetal force, as indicated in this equation.
5.1.3
Newton’s Law of Gravity (Optional for Calculus–Based)
The force of gravity is one of the fundamental forces in nature. Although in our first physics
examples we only dealt with the fact that the earth pulls downward on all masses, in fact all
masses exert an attractive gravitational force on each other, but for most objects the force
is so small that we can ignore it.
Newton’s Law of Gravity says that for two masses m1 and m2 separated by a distance
r, the magnitude of the (attractive) gravitational force is
F =G
m1 m2
r2
where G = 6.67 × 10−11
N·m2
kg2
(5.4)
While the law as given really applies to point (i.e. small) masses, it can be used for spherical
masses as long as we take r to be the distance between the centers of the two masses.
5.2
5.2.1
Worked Examples
Friction Forces
1. An ice skater moving at 12 ms coasts to a halt in 95 m on an ice surface. What
is the coefficient of (kinetic) friction between ice and skates? [Ser4 5-51]
The information which we are given about the skater’s (one-dimensional) motion is shown
in Fig. 5.1(a). We know that the skater’s notion is one of constant acceleration so we can use
the results in Chapter 2. In particular we know the initial and final velocities of the skater:
v0 = 12 ms
v=0
and we know the displacement for this interval:
x − x0 = 95 m
we can use 2.8 to find the (constant) acceleration a. We find:
2
vx2 = v0x
+ 2ax (x − x0 )
=⇒
ax =
2
(vx2 − v0x
)
2(x − x0)
105
5.2. WORKED EXAMPLES
N
x
x
v=12 m/s
v=0
fk
95 m
mg
(a)
(b)
Figure 5.1: Skater slowed to a halt by friction in Example 1. Motion is shown in (a); forces acting on the
skater are shown in (b).
Substituting, we get:
ax =
((0 ms )2 − (12 ms )2 )
= −0.76 sm2 .
2(95 m)
Next, think about the forces acting on the skater; these are shown in Fig. 5.1(b). If the
mass of the skater is m then gravity has magnitude mg and points downward; the ice exerts
a normal force N upward. It also exerts a frictional force fk in a direction opposing the
motion. Since the skater has no motion in the vertical direction, the vertical forces must
sum to zero so that N = mg. Also, since the magnitude of the force of kinetic friction is
given by fk = µk N we have:
fk = µk N = µk mg .
So the net force in the x direction is Fx = −µk mg.
Newton’s law tells us: Fx, net = max. Using the results we have found, this gives us:
−µk mg = m(−0.76 sm2 )
From which the m cancels to give:
µk =
(0.76 m
)
(0.76 sm2 )
s2
=
= 7.7 × 10−2
g
(9.80 sm2 )
The coefficient of kinetic friction between ice and skates is 7.7 × 10−2 . (Note, the coefficient
of friction is dimensionless.)
Recall that we were not given the mass of the skater. That didn’t matter, because it
cancelled out of our equations. But we did have to consider it in writing down our equations.
2. Block B in Fig. 5.2 weighs 711 N. The coefficient of static friction between
block and table is 0.25; assume that the cord between B and the knot is horizontal.
Find the maximum weight of block A for which the system will be stationary.
[HRW6 6-19]
We need to look at the forces acting at the knot (the junction of the three cables). These
are shown in Fig. 5.3(a). The vertical cord must have a tension equal to the weight of block
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CHAPTER 5. FORCES AND MOTION II
30o
B
A
Figure 5.2: Diagram for Example 2.
N
T2
B
T1
T1
fs
WA
WB
(a)
(b)
Figure 5.3: (a) Forces acting at the knot in Example 2. (b) Forces acting on block B in Example 2.
A (which we’ll call WA ) because at its other end this cord is pulling up on A so as to support
it. Let the tensions in the other cords be T1 for the horizontal one and T2 for the one that
pulls at 30◦ above the horizontal. The knot is in equilibrium so the forces acting on it add
to zero. In particular, the vertical components of the forces add to zero, giving:
T2 sin θ − WA = 0
or
T2 sin θ = WA
(where θ = 30◦ ) and the horizontal forces add to zero, giving:
−T1 + T2 cos θ = 0
or
T1 = T2 cos θ .
Now look at the forces acting on the block which rests on the table; these are shown in
Fig. 5.3(b). There is the force of gravity pointing down, with magnitude WB (that is, the
weight of B, equal to mB g). There is a normal force from the table pointing upward; there
is the force from the cable pointing to the right with magnitude T1 , and there is the force of
static friction pointing to the left with magnitude fs . Since the vertical forces add to zero,
we have
N − WB = 0
or
N = Wb
The horizontal forces on the block also sum to zero giving
T1 − fs = 0
or
T1 = fs
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5.2. WORKED EXAMPLES
m
F
M
Frictionless
Figure 5.4: Diagram for Example 3.
Now, the problem states that the value of WA we’re finding is the maximum value such
that the system is stationary. This means that at the value of WA we’re finding, block B is
just about to slip, so that the friction force fs takes on its maximum value, fs = µs N. Since
we also know that N = WB from the previous equation, we get:
T1 = fs = µs N = µs WB
From before, we had T1 = T2 cos θ, so making this substitution we get
T2 cos θ = µs WB
Almost done! Our very first equation gave T2 =
WA
cos θ = µs WB
sin θ
or
WA
,
sin θ
so substituting for T2 gives:
WA cot θ = µs WB
Finally, we get:
WA = µs WB tan θ
Now just plug in the numbers:
WA = (0.25)(711 N) tan 30◦ = 103 N
Since we solved for WA under the condition that block B was about to slip, this is the
maximum possible value for WA so that the system is stationary.
3. The two blocks (with m = 16 kg and M = 88 kg) shown in Fig. 5.4 are not
attached. The coefficient of static friction between the blocks is µs = 0.38, but
the surface beneath M is frictionless. What is the minimum magnitude of the
horizontal force F required to hold m against M? [HRW5 6-38]
Having understood the basic set-up of the problem, we immediately begin thinking about
the the forces acting on each mass so that we can draw free–body diagrams. The forces on
mass m are: (1) The force of gravity mg which points downward. (2) The applied force F
which points to the right. (3) The normal force with which block M pushes on m. This
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CHAPTER 5. FORCES AND MOTION II
Nsurf
x
fs
F
N
N
mg
Mg
fs
(a)
(b)
Figure 5.5: Free–body diagrams for the blocks described in Example 3.
force necessarily points to the left. (4) The frictional force which block M exerts on m. This
is to be a static friction force, so we have to think about its direction... in this case, it must
clearly oppose the force of gravity to keep the block m from falling. So we include a force fs
pointing up. These forces are shown in Fig. 5.5.
Next we diagram the forces acting on M. There is the force of gravity, with magnitude
Mg, pointing down; the surface beneath M exerts a normal force N pointing upward. Since
this surface is frictionless, it does not exert a horizontal force on M. The mass m will exert
forces on M and these will be equal and opposite to the forces which M exerts on m. So there
is a force N on mass M pointing to the right and a frictional force fs pointing downward.
Now that we have shown all the forces acting on all the masses we can start to discuss
the accelerations of the masses and apply Newton’s Second Law.
The problem says that mass m is not slipping downward during its motion. This must
mean that the forces of friction and gravity balance:
fs = mg .
But this does us little good until we have an expression for fs . Now, in this problem we are
being asked about a critical condition for the slippage of m. We can reasonably guess that
here the force of static friction takes on its maximum value, namely
fs = µs N ,
N being the normal force between the two surfaces. This is an important bit of information,
because combining that last two equations we get:
mg = µs N .
Let’s consider the horizontal motion of both of the masses. Now, since the masses are
always touching, their displacements, velocities and accelerations are always the same. Let
the x acceleration of both masses be a. Then for mass m, Newton’s Second Law gives us:
X
Fx = F − N = ma
109
5.2. WORKED EXAMPLES
m1
m2
q
Figure 5.6: Two blocks joined by a rod slide down a slope with friction (coefficient of friction is different
for the two blocks).
while for mass M, we get
N = Ma
Combining these last two equations gives
F − Ma = ma
=⇒
F = (M + m)a
which tells us the force N:
N = Ma =
=⇒
a=
F
(M + m)
MF
(M + m)
Putting this result for N into our result involving the friction force gives
mg = µs N = µs
which lets us solve for F :
F =
MF
(M + m)
(M + m)m
g
Mµs
And now we can substitute the given values:
F =
(M + m)m
(16 kg + 88 kg)(16 kg)
g=
(9.80 sm2 ) = 488 N
Mµs
(88 kg)(0.38)
4. In Fig. 5.6 a box of mass m1 = 1.65 kg and a box of mass m2 = 3.30 kg slide
down an inclined plane while attached by a massless rod parallel to the plane.
The angle of incline is θ = 30◦ . The coefficient of kinetic friction between m1 and
the incline is µ1 = 0.226; that between m2 and the incline is µ2 = 0.113. Compute
(a) the tension in the rod and (b) the common acceleration of the two boxes. c)
How would the answers to (a) and (b) change if m2 trailed m1 ? [HRW6 6-26]
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CHAPTER 5. FORCES AND MOTION II
m2
m1
N1
fk, 1
fk, 2
N2
T
T
m1g sin q
m1g cos q
m2g sin q
m2g cos q
m1g
(a)
m2g
(b)
Figure 5.7: (a) Forces acting on block 1 in Example 4. We have assumed that the rod pushes outward; if
that is wrong, then T will turn out to be negative. The force of gravity has be split up into components.
(b) Forces acting on block 2 in Example 4.
(a) We will shortly be drawing force diagrams for the two masses, but we should first pause
and consider the force which comes from the rod joining the two masses. A “rod” differs
from a “cord” in our problems in that it can pull inward on either end or else push outward .
(Strings can only pull inward.) For the purpose of writing down our equations we need to
make some assumption about what is happening and so here I will assume that the rod is
pushing outward with a force of magnitude T , i.e. the rod is compressed. Should it arise in
our solution that we get a negative number for T , all is not lost; we will then know that the
rod is really pulling inward with a force of magnitude |T | and so the rod is being stretched.
With that in mind, we draw a diagram for the forces acting on block 1 and there are a
lot of them, as shown in Fig. 5.7(a). Rod tension T and the force of kinetic friction on block
1 (to oppose the motion) point up the slope. The “slope” component of gravity m1g sin θ
points down the slope. The normal component of gravity m1g cos θ points into the surface
and the normal force N1 from the slope points out of the surface.
As there is no acceleration perpendicular to the slope, those force components sum to
zero, giving:
N1 − m1g cos θ = 0
or
N1 = m1 g cos θ
The sum of force components in the down–the–slope direction gives m1 a, where a is the
down–the–slope acceleration common to both masses. So then:
m1 g sin θ − T − fk,1 = m1a
We can substitute for fk,1 , since fk,1 = µ1 N1 = µ1 m1 g cos θ. That gives:
m1 g sin θ − T − µ1 m1g cos θ = m1a
(5.5)
We have a fine equation here, but T and a are both unknown; we need another equation!
The forces acting on block 2 are shown in Fig. 5.7(b). The force of kinetic friction fk,2
points up the slope. The rod tension T and the “slope” component of gravity m2 g sin θ point
111
5.2. WORKED EXAMPLES
down the slope. The normal component of gravity m2g cos θ points into the surface and the
normal force of the surface on 2, N2 , points out of the slope.
Again there is no net force perpendicular to the slope, so
N2 − m2 g cos θ = 0
or
N2 = m2g cos θ .
The sum of the down–the–slope forces on m2 gives m2a, so:
m2g sin θ + T − fk,2 = m2 a
We can substitute for the force of kinetic friction here, with fk,2 = µ2 N2 = µ2 m2g cos θ.
Then:
m2g sin θ + T − µ2 m2 g cos θ = m2 a
(5.6)
Two equations (5.5 and 5.5) and tow unknowns (T and a). The physics is done, the rest
is math!
In solving the equations I will go for an analytic (algebraic) solution, then plug in the
numbers at the end. Aside from giving us some good practice with algebra, it will be useful
in answering part (c).
We note that if we add Eqs. 5.5 and 5.5, T will be eliminated and we can then find a.
When we do this, we get:
m1 g sin θ + m2 g sin θ − µ1 g cos θ − µ2 g cos θ = ma + m2a
Lots of factoring to do here! Pulling out some common factors, this is:
g [(m1 + m2) sin θ − cos θ(µ1 m1 + µ2 m2)] = (m1 + m2)a
and then we get a:
a=
g [(m1 + m2 ) sin θ − cos θ(µ1 m1 + µ2 m2 )]
(m1 + m2)
(5.7)
But it’s really T we want in part (a). We can eliminate a by multiplying Eq. 5.5 by m2:
m1m2g sin θ − m2T − µ1 m1m2g cos θ = m1m2 a
and Eq. 5.6 by m2:
m1m2 g sin θ + m1 T − µ2 m1 m2g cos θ = m1m2a
and then subtracting the second from the first. Some terms cancel, and this gives:
−m2T − m1 T − µ1 m1m2 g cos θ + µ2 m1m2 g cos θ = 0
Factor things:
−T (m1 + m2) = m1m2 g cos θ(µ1 − µ2 )
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CHAPTER 5. FORCES AND MOTION II
and finally get an expression for T :
T =
m1m2 g cos θ(µ2 − µ1 )
(m1 + m2)
(5.8)
Hey, that algebra wasn’t so bad, was it? Now we have general expressions for T and a.
Plugging numbers into Eq. 5.8, we get:
(1.65 kg)(3.30 kg)(9.80 sm2 ) cos 30◦ (0.113 − 0.226)
(1.65 kg + 3.30 kg)
= −1.05 N
T =
Oops! T came out negative! What we find from this is that the assumption was wrong and
the rod is really being stretched as the blocks slide down the slope, and the magnitude of
the rod’s tension is 1.05 N.
(b) To find the acceleration of the blocks, plug numbers into Eq. 5.7:
(9.80 sm2 ) [(1.65 kg + 3.30 kg) sin 30◦ − cos 30◦ ((0.226)(1.65 kg) + (0.113)(3.30 kg)]
(1.65 kg + 3.30 kg)
m
= 3.62 s2
a =
The (common) acceleration of the blocks is 3.62 sm2 .
(c) Now we ask: What would the answers to (a) and (b) be if the blocks had slid down the
slope with m1 in the lead? Would it make any difference at all? It might, since the friction
forces on the masses come from two different µ’s. In any case, with our analytic results
Eqs. 5.7 and 5.8 we can find the results of switching the labels “1” and “2”, since that is all
we would get from having the blocks switch positions.
If we switch “1” and “2” in Eq. 5.7, we can see that the result for a will not change at all
because the sums within that expression are not affected by the switch. So the connected
blocks will slide down the slope with the same acceleration, namely 3.62 sm2 for the given
values.
What about T ? From Eq. 5.8 we see that switching “1” and “2” gives an overall change in
sign because of the factor (µ2 − µ1 ). (The other factors don’t change for this switch.) So we
know that plugging in the numbers for the case where blocks 1 leads would give T = +1.05 N,
and since this is a positive number, the assumption about the rod being compressed (and as
a result pushing outward) would be correct. So for the case where m1 leads, the magnitude
of the rod’s tension is the same (1.05 N) , but now it pushing outward .
5. A 3.0 − kg block starts from rest at the top of a 30.0◦ incline and slides 2.0 m
down the incline in 1.5 s. Find (a) the magnitude of the acceleration of the block,
(b) the coefficient of kinetic friction between the block and the plane, (c) the
frictional force acting on the block and (d) the speed of the block after it has
slid 2.0 m. [Ser4 5-57]
113
5.2. WORKED EXAMPLES
y
y
t=0
fk
N
x
x
2.0 m
v=0
mg sin q
t = 1.5 s
30o
q
mg cos q
mg
(a)
(b)
Figure 5.8: (a) Block slides down rough slope in Example 5. (b) Forces acting on the block.
(a) The basic information about the motion of the block is summarized in Fig. 5.8(a). We
use a coordinate system where x points down the slope and y is perpendicular to the slope.
We’ll put the origin of the coordinate system at the place where the block begins its motion.
The block’s motion down the slope is one of constant acceleration. (This must be so,
since all of the forces acting on the block are constant.) Of course, this is an acceleration in
the x direction, as there is no y motion. It begins its slide starting from rest (v0x = 0) and
so the block’s motion is given by:
x = x0 + v0xt + 21 ax t2 = 12 ax t2 .
We are told that at t = 1.5 s, x = 2.0 m. Substitute and solve for ax :
2.0 m = 21 ax (1.5 s)2
=⇒
ax =
2(2.0 m)
= 1.78 sm2
(1.5 s)2
The magnitude of the block’s acceleration is 1.78 sm2 .
(b) We must now think about the forces which act on the block. They are shown in
Fig. 5.8(b). Gravity pulls downward with a force mg, which we decompose into its components along the slope and perpendicular to it. The surface exerts a normal force N. There
is also a force of kinetic friction from the slope. Since the block is moving down the slope,
the friction force must point up the slope.
The block moves only along the x axis; the forces in the y direction must sum to zero.
Referring to Fig. 5.8(b), we get:
X
Fy = N − mg cos θ = 0
=⇒
N = mg cos θ .
This gives us the normal force of the surface on the block; here, θ = 30.0◦ .
The block does have an acceleration in the x direction, which we’ve already found in part
(a). The sum of the forces in the +x direction gives max:
X
Fx = mg sin θ − fk = max
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CHAPTER 5. FORCES AND MOTION II
Now we use the formula for the force of kinetic friction: fk = µk N. Using our expression for
the normal force gives us:
fk = µk N = µk mg cos θ
and using this result in the last equation gives
mg sin θ − µk mg cos θ = max .
Here, the only unknown is µk , so we find it with a little algebra: First off, we can cancel the
common factor of m that appears in all terms:
g sin θ − µk g cos θ = ax
and then solve for µk :
µk g cos θ = g sin θ − ax
= (9.80 sm2 ) sin 30.0◦ − (1.78 sm2 ) = 3.12 sm2
So we get:
µk =
(3.12 sm2 )
= 0.368
(9.80 sm2 )(cos 30.0◦ )
(c) As we have seen in part (b), the magnitude of the (kinetic) friction force on the mass is
fk = µk mg cos θ
= (0.368)(3.0 kg)(9.80 sm2 ) cos 30.0◦
= 9.4 N
The force of friction is 9.4 N.
(d) We know the acceleration of the block, its initial velocity (v0x = 0) and the time of travel
to slide 2.0 m; its final velocity is
v = v0x + ax t = 0 + (1.78 m
)(1.50 s) = 2.67 ms
s2
6. Three masses are connected on a table as shown in Fig. 5.9. The table has a
coefficient of sliding friction of 0.35. The three masses are 4.0 kg, 1.0 kg, and 2.0 kg,
respectively and the pulleys are frictionless. (a) Determine the acceleration of
each block and their directions. (b) Determine the tensions in the two cords.
[Ser4 5-59]
(a) First, a little thinking about what we expect to happen. Surely, since the larger mass is
hanging on the left side we expect the 4.0 kg mass to accelerate downward, the 1.0 [kg block
to accelerate to the right and the 2.0 kg block to accelerate upward. Also, since the masses
are connected by strings as shown in the figure, the magnitudes of all three accelerations
115
5.2. WORKED EXAMPLES
mk = 0.35
1.0 kg
2.0 kg
4.0 kg
Figure 5.9: System for Example 6
T2
T1
N
T2
T1
fk
m2 g
m3g
m1g
(a)
(b)
(c)
Figure 5.10: Free–body diagrams for the three masses in Example 6. (a) Forces on the mass m1 = 4.0 kg.
(b) Forces on the mass m2 = 1.0 kg. (c) Forces on the mass m3 = 2.0 kg. The directions of motion assumed
for each mass are also shown.
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CHAPTER 5. FORCES AND MOTION II
must be the same, because in any time interval the magnitudes of their displacements will
always be the same. So each mass will have an acceleration of magnitude a with the direction
appropriate for each mass.
Now we consider the forces acting on each mass. We draw free–body diagrams! If the
tension in the left string is T1 then the forces on the 4.0 kg mass are as shown in Fig. 5.10(a).
The string tension T1 pulls upward; gravity pulls downward with a force m1g.
The forces acting on m2 are shown in Fig. 5.10(b). We have more of them to think about;
gravity pulls with a force m2g downward. The table pushes upward with a normal force N.
It also exerts a frictional force on m2 which opposes its motion. Since we think we know
which way m2 is going to go (left!), the friction force fk must point to the right. There are
also forces from the strings. There is a force T1 to the left from the tension in the first string
and a force T2 pointing to the right from the tension in the other string. (Note, since these
are two different pieces of string, they can have different tensions.)
The forces on m3 are shown in Fig. 5.10(c). There is a string tension T2 pulling up and
gravity m3 g pulling down.
All right, lets write down some equations! By Newton’s Second Law, the sum of the downward forces on m1 should give m1 a. (We agreed that its acceleration would be downward.)
This gives:
m1 g − T1 = m1 a
(5.9)
Moving on to mass m2, the vertical forces must cancel, giving
N = m2 g .
Newton tells us that the sum of the left–pointing forces must give m2a (we decided that its
acceleration would be of magnitude a, toward the left) and this gives:
T1 − fk − T2 = m2a
But since
fk = µk N = µk m2 g ,
this becomes
T1 − µk m2g − T2 = m2a .
(5.10)
Finally, the sum of the upward forces on m3 must give m3a. So:
T2 − m3 g = m3 a
(5.11)
Having done this work in writing down these wonderful equations we stand back, admire
our work and ask if we can go on to solve them. We note that there are three unknowns (a,
T1 and T2) and we have three equations. We can find a solution. The physics is done. . . only
the algebra remains.
We can do the algebra in the following way: If we just add Eqs. 5.9, 5.10 and 5.11 together
(that is, add all the left–hand–sides together and the right–hand–sides together) we find that
both T ’s cancel out. We get:
m1 g − T1 + T1 − µk m2g − T2 +2 −m3g = m1 a + m2a + m3a
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5.2. WORKED EXAMPLES
which simplifies to:
m1 g − µk m2 g − m3g = (m1 + m2 + m3)a
Now we can easily find a:
(m1 − µk m2 − m3)g
(m1 + m2 + m3)
[(4.0 kg) − (0.35)(1.0 kg) − (2.0 kg)](9.80 sm2 )
=
(4.0 kg + 1.0 kg + 2.0 kg)
(1.65 kg)(9.80 sm2 )
=
= 2.31 sm2
(7.0 kg)
a =
So our complete answer to part (a) is: m1 accelerates at 2.31 sm2 downward; m2 accelerates
at 2.31 sm2 to the left; m3 accelerates at 2.31 sm2 upward.
(b) Finding the tensions in the strings is now easy; just use the equations we found in part
(a).
To get T1 , we can use Eq. 5.9, which gives us:
T1 = m1 g − m1a = m1 (g − a) = (4.0 kg)(9.80 sm2 − 2.31 sm2 ) = 30.0 N .
To get T2 we can use Eq. 5.11 which gives us:
T2 = m3 g + m3a = m3(g + a) = (2.0 kg)(9.80 sm2 + 2.31 sm2 ) = 24.2 N .
The tension in the string on the left is 30.0 N. The tension in the string on the right is
24.2 N.
7. A block is placed on a plane inclined at 35◦ relative to the horizontal. If the
block slides down the plane with an acceleration of magnitude g/3, determine
the coefficient of kinetic friction between block and plane. [Ser4 5-61]
The forces acting on the block (which has mass m) as it slides down the inclined plane are
shown in Fig. 5.11. The force of gravity has magnitude mg and points straight down; here
it is split into components normal to the slope and down the slope, which have magnitudes
mg cos θ and mg sin θ, respectively, with θ = 35◦ . The surface exerts a normal force N and
a force of kinetic friction, fk , which, since the block is moving down the slope, points up the
slope.
The block can only accelerate along the direction of the slope, so the forces perpendicular
to the slope must add to zero. This gives us:
N − mg cos θ = 0
=⇒
N = mg cos θ
The acceleration of the block down the slope was given to us as a = g/3. Then summing
the forces which point along the slope, we have
mg sin θ − fk = ma = mg/3
118
CHAPTER 5. FORCES AND MOTION II
N
fk
Dir. of motion
mg sin q
mg cos q
350
mg
Figure 5.11: Forces on the block in Example 7.
The force of kinetic friction is equal to µk N, and using our expression for N we have
fk = µk N = µk mg cos θ
and putting this into the previous equation gives:
mg sin θ − µk mg cos θ = mg/3 .
Fortunately, the mass m cancels from this equation; we get:
g sin θ − µk g cos θ = g/3
And now the only unknown is µk which we solve for:
µk g cos θ = g sin θ −
g
= g(sin θ − 31 )
3
Here we see that g also cancels, although we always knew the value of g! We then get:
(sin θ − 13 )
(sin 35◦ − 13 )
µk =
=
= 0.293
cos θ
cos 35◦
So the coefficient of kinetic friction between block and slope is 0.293.
8. A 2.0 kg block is placed on top of a 5.0 kg as shown in Fig. 5.12. The coefficient
of kinetic friction between the 5.0 kg block and the surface is 0.20. A horizontal
force F is applied to the 5.0 kg block. (a) Draw a free–body diagram for each
block. What force accelerates the 2.0 kg block? (b) Calculate the magnitude of
the force necessary to pull both blocks to the right with an acceleration of 3.0 sm2 .
(c) find the minimum coefficient of static friction between the blocks such that
the 2.0 kg block does not slip under an acceleration of 3.0 sm2 . [Ser4 5-73]
(a) What forces act on each block?
On the big block (with mass M = 5.0 kg, let’s say) we have the applied force F which
pulls to the right. There is the force of gravity, Mg downward. The surface exerts a normal
119
5.2. WORKED EXAMPLES
2.0 kg
mk = 0.20
F
5.0 kg
Figure 5.12: Figure for Example 8.
M
N2
m
N1
fs
fs
F
fk
N2
Mg
(a)
mg
(b)
Figure 5.13: (a) Forces acting on the large block, M = 5.0 kg. (b) Forces acting on the small block,
m = 2.0 kg.
force N1 upward. There is a friction force from the surface, which is directed to the left. The
small mass will also exert forces on mass M; it exerts a normal force N2 which is directed
downward ; we know this because M is pushing upward on m. Now, M is exerting a force
of static friction fs on m which goes to the right; so m must exert a friction force fs on M
which points to the left.
These forces are diagrammed in Fig. 5.13(a).
On the small block we have the force of gravity, mg downward. Mass M exerts an upward
normal force N2 on it, and also a force of static friction fs on it, pointing to the right. It
is this force which accelerates m as it moves along with M (without slipping). These forces
are diagrammed in Fig. 5.13(b).
Notice how the forces between M and m, namely N2 (normal) and fs , have the same
magnitude but opposite directions, in accordance with Newton’s Third Law. They are so–
called “action–reaction pairs”.
(b) The blocks will have a horizontal acceleration but no vertical motion, so that allows us
to solve for some of the forces explained in part (a). The vertical forces on m must sum to
zero, giving us:
N2 − mg = 0
=⇒
N2 = mg = (2.0 kg)(9.80 sm2 ) = 19.6 N
120
CHAPTER 5. FORCES AND MOTION II
and the vertical forces on M must sum to zero, giving us:
N1 − N2 − Mg = 0
=⇒
N1 = N2 + Mg = 19.6 N + (5.0 kg)(9.80 m
) = 68.6 N
s2
We are given that the acceleration of both blocks is 3.0 sm2 . Applying Newton’s Second
Law to mass m we find:
X
Fx = fs = max = (2.0 kg)(3.0 sm2 ) = 6.0 N
While applying it to M gives
X
Fx = F − fk − fs = Max = (5.0 kg)(3.0 sm2 ) = 15.0 N
We found fs above; we do know the force of kinetic friction (from M’s sliding on the surface)
because we know the coefficient of kinetic friction and the normal force N1:
fk = µk N1 = (0.20)(68.6 N) = 13.7 N
Now we can solve for F :
F = 15.0 N + fk + fs
= 15.0 N + 13.7 N + 6.0 N
= 34.7 N
To pull the blocks together to the right with an acceleration 3.0 sm2 we need an applied force
of 34.7 N.
(c) As we’ve seen, mass m accelerates because of the friction force fs (from M’s surface)
which acts on it. Forces of static friction have a maximum value; here we know that we must
have
fs ≤ µs N2
in order for m not to slip on M. Here, we have fs = 6.0 N and N2 = 19.6 N. So the critical
value of µs for our example is
f2
µs =
= 0.306
N2
If µs is less than this value, static friction cannot supply the force needed to accelerate m
at 3.0 sm2 . So µs = 0.306 is the minimum value of the coefficient of static friction so that the
upper block does not slip.
5.2.2
Uniform Circular Motion Revisited
9. A toy car moving at constant speed completes one lap around a circular track
(a distance of 200 m) in 25.0 s. (a) What is the average speed? (b) If the mass of
the car is 1.50 kg, what is the magnitude of the central force that keeps it in a
circle? [Ser4 6-1]
121
5.2. WORKED EXAMPLES
r
m
M
Figure 5.14: Mass m moves; mass M hangs!
(a) If a lap around the circular track is of length 200 m then the (average) speed of the car
is
200 m
d
= 8.00 ms
v= =
t
25.0 s
(b) The car undergoes uniform circular motion, moving in a circle of radius r with speed v.
The net force on the car points toward the center of the circle and has magnitude
mv 2
Fcent =
r
Actually, we haven’t found r yet. We are given the circumference of the circle, and from
C = 2πr we find
C
200 m
=
= 31.8 m
r=
2π
2π
So the net force on the car has magnitude
Fcent
(1.50 kg)(8.00 ms )2
mv 2
=
=
= 3.02 N
r
(31.8 m)
The net force on the car has magnitude 3.02 N; its direction is always inward, keeping the
car on a circular path.
10. A mass M on a frictionless table is attached to a hanging mass M by a cord
through a hole in the table, as shown in Fig. 5.14. Find the speed with which
m must move in order for M to stay at rest. [HRW5 6-57]
Taking mass M to be at rest, we see that mass m must be moving in a circle of constant
radius r. It is moving at (constant) speed v; so mass m undergoes uniform circular motion.
So the net force on m points toward the center of the circle and has magnitude Fcent = mv 2/r.
The free–body diagram for m is shown in Fig. 5.15(a). The only force on m is the string
122
CHAPTER 5. FORCES AND MOTION II
T
T
acent
Mg
(b)
(a)
Figure 5.15: (a) Force on mass m and the direction of its acceleration. (There are also vertical forces,
gravity and the table’s normal force, which cancel; these are not shown.) (b) Forces acting on hanging mass
M.
tension (pointing toward the center of the circle). This gives us:
T =
mv 2
r
Next consider the forces acting on M and its motion. The force diagram for M is shown
in Fig. 5.15(b). Since mass M is at rest, the net force on it is zero, which gives:
T = Mg
Combining these two results, we get:
mv 2
= Mg
r
Solving for v, we get:
Mgr
v2 =
m
=⇒
v=
s
Mgr
m
11. A stuntman drives a car over the top of a hill, the cross section of which can
be approximated by a circle of radius 250 m, as in Fig. 5.16. What is the greatest
speed at which he can drive without the car leaving the road at the top of the
hill? [HRW5 6-58]
We begin by thinking about the forces acting on the car and its acceleration when it is
at the top of the hill.
At the top of the hill, the car is moving in a circular path of radius r = 250 m with some
speed v. Then the car has a centripetal acceleration of magnitude v 2/r which is directed
downward. (For all we know, it may also have a tangential acceleration as well, but the
123
5.2. WORKED EXAMPLES
250 m
Figure 5.16: Car drives over the top of a hill in Example 11.
N
v
mg
Figure 5.17: Forces acting on the car in Example 11 when it is at the top of the hill.
problem gives no information on it, and it won’t be relevant for the problem.) By Newton’s
Second Law, the net (vertical) force on the car must have magnitude mv 2/r and must be
directed downward .
The forces acting on the car are shown in Fig. 5.17. Then the force of gravity is mg
downward. The road exerts a normal force of magnitude N upward. One may ask how we
know the road’s force goes upward. This is because there is no physical way in which a road
can pull downward on a car driving over it. But it can push up.
We combine the results from the last two paragraphs. The net downward force must
equal mv 2/r. This gives us:
mv 2
.
mg − N =
r
however without knowing anything more, we can’t solve for v in this equation because we
don’t know N (or, for that matter, m).
We have not yet used the condition that the car is on the verge of leaving the road at
the top of the hill. What does this condition give us? If we use the last equation to find the
normal force:
mv 2
N = mg −
r
we see that if we increase v there comes a point at which N must be negative in order for
the car to stay on the road moving on its circular arc. But as discussed above, N can’t be
negative. But it can be zero, and it is for this speed that the car is on the verge of leaving
the road at the top of the hill. The critical case has N = 0, and this gives us:
0 = mg −
mv 2
r
=⇒
mv 2
= mg .
r
124
CHAPTER 5. FORCES AND MOTION II
v
r
a
Figure 5.18: Coin moves with a rotating turntable
Toward center
N
fs
mg
Figure 5.19: Forces acting on the coin in Example 12
Note that the mass m cancels out of this equation so we don’t need to know m. We get:
v 2 = rg = (250 m)(9.80 sm2 ) = 2.45 × 103
m2
s2
and finally
v = 49.5 ms .
The car may be driven as fast as 49.5 ms and it will stay on the road.
12. A coin placed 30.0 cm from the center of a rotating, horizontal turntable slips
when its speed is 50.0 cm
. (a) What provides the central force when the coin is
s
stationary relative to the turntable? (b) What is the coefficient of static friction
between the coin and turntable? [Ser4 6-13]
(a) See Fig. 5.18 for a fine illustration of the problem.
As the coin executes uniform circular motion (before it slips) it is accelerating toward
the center of the turntable! So there must be a force (or forces) on the coin causing it to
do this. This force can only come from its contact interaction with the turntable, i.e. from
friction. Here, since we are dealing with the case where the coin is not sliding with respect
to the surface, it is the force of static friction. Furthermore, the force of static friction is
directed toward the center of the turntable.
(b) A view of the forces acting on the coin is given in Fig. 5.19. If the mass of the coin is
m then gravity exerts a force mg downward, the turntable exerts a normal force N upward
125
5.2. WORKED EXAMPLES
and there is a force of static friction, which as we discussed in part (a) must point toward
the center of the turntable.
The acceleration of the coin points toward the center of the circle and has magnitude v 2/r,
(r being the distance of the coin from the center). So the vertical forces must cancel, giving
us N = mg. The net force points inward and has magnitude mv 2/r, so that fs = mv 2/r.
Now for the conditions at which the coin starts to slip, the force of static friction has
reached its maximum value, i.e.
fs = µs N
but here we can use our results to substitute for fs and for N. This give us:
mv 2
= µs mg
r
which lets us solve for µ:
(50.0 cm
)2
(0.500 ms )2
v2
−2
s
µs =
=
m =
m = 8.50 × 10
rg
(30.0 cm)(9.80 s2 )
(0.300 m)(9.80 s2 )
So the coefficient of static friction for the turntable and coin is µs = 8.50 × 10−2 .
We were never given the mass of the coin, but we did not need it because it cancelled
out of our equations just before the final answer.
13. A Ferris wheel rotates four times each minute; it has a diameter of 18.0 m.
(a) What is the centripetal acceleration of a rider? What force does the seat
exert on a 40.0 − kg rider (b) at the lowest point of the ride and (c) at the highest
point of the ride? (d) What force (magnitude and direction) does the seat exert
on a rider when the rider is halfway between top and bottom? [Ser4 6-51]
(a) First, calculate some numbers which we know are important for circular motion! The
wheel turns around 4 times in one minute, so the time for one turn must be
T =
1.0 min
60.0 s
=
= 15 s .
4
4
Also, since the radius of the wheel is R = D/2 = 18.0 m/2 = 9.0 m, the circumference of the
wheel is
C = 2πR = 2π(9.0 m) = 57 m
and then the speed of a rider is
v=
C
57 m
=
= 3.8 ms .
T
15 s
The rider moves at constant speed in a circular path of radius R. So the rider’s acceleration
is always directed toward the center of the circle and it has magnitude
acent =
(3.8 ms )2
v2
=
= 1.6 sm2
R
(9.0 m)
126
CHAPTER 5. FORCES AND MOTION II
Fseat
Fseat
a
a
Mg
Mg
(b)
(a)
Figure 5.20: Forces acting on the Ferris wheel rider (a) at the lowest point of the ride and (b) at the
highest point of the ride.
The centripetal acceleration of the rider is 1.6 sm2 .
(b) Consider what is happening when rider is at the lowest point of the ride. His acceleration
is upward (toward the center of the circle!) and has magnitude 1.6 sm2 .
What are the forces acting on the rider (who has mass M, let’s say) at this point? These
are shown in Fig. 5.20(a). Gravity pulls down on the rider with a force of magnitude Mg, and
the seat pushes upward on the rider with a force Fseat. (Usually seats can’t pull downward;
also, the force of the seat can’t have any sideways component because here the net force
must point upward ). Since the net force points upward and has magnitude Fcent = Mv 2 /R,
Newton’s Second Law gives us:
Mv 2
Fseat − Mg =
R
Since M = 40.0 kg, we get:
Fseat
(40.0 kg)(3.8 ms )2
Mv 2
m
= (40.0 kg)(9.80 s2 ) +
= 460 N
= Mg +
R
(9.0 m)
The seat pushes upward on the rider with a force of magnitude 460 N. We might say that
when the rider at the lowest point, the rider has an apparent weight of 460 N, since that is
the force of the surface on which the rider rests. Here, the apparent weight is greater than
the true weight Mg of the rider.
(c) When the rider is at the highest point of the wheel, his acceleration is downward . The
forces acting on the rider are shown in Fig. 5.20(b); these are the same forces as in part (a)
but now the net force points downward and has magnitude Fcent = Mv 2/R. Adding up the
downward forces, Newton’s Second Law now gives us:
Mg − Fseat =
Mv 2
R
which now gives us
Fseat = Mg −
(40.0 kg)(3.8 ms )2
Mv 2
= (40.0 kg)(9.80 sm2 ) −
= 330 N .
R
(9.0 m)
127
5.2. WORKED EXAMPLES
Fseat
a
Mg
Figure 5.21: Forces on the rider when he is halfway between top and bottom.
The seat pushes upward on the rider with a force of magnitude 330 N. We would say that at
the top of the ride, the apparent weight of the rider is 330 N. This time the apparent weight
is less than the true weight of the rider.
(d) When the rider is halfway between top and bottom, the net force on him still points
toward the center of the circle (and has magnitude Mv 2/R), but in this case the direction
is horizontal, as indicated in Fig. 5.21. (In this picture the rider is on the right side of the
Ferris wheel, as we look at it face–on.) The forces acting on the rider are also shown in this
picture. The force of gravity, Mg can only pull downward. The only other force on the rider,
namely that of the seat does not push straight upward in this case. We know that it can’t,
because the sum of the two forces must point horizontally (to the right). The force of the
seat must also have a horizontal component; it must point as shown in Fig. 5.21.
Without being overly formal about the mathematics we can see that the vertical component of Fseat must be Mg so as to cancel the force of gravity. The vertical component of
Fseat must have magnitude
Fseat, vert = Mg = (40.0 kg)(9.80 sm2 ) = 392 N
The horizontal component of Fseat must equal Mv 2 /R since as we’ve seen, that is the net
force on the rider. So:
(40.0 kg)(3.8 ms )2
Mv 2
Fseat, horiz =
=
= 64 N
R
(9.0 m)
Now we can find the magnitude of the force of the seat:
Fseat =
q
Fseat, vert2 + Fseat, horiz2
q
=
(392 N)2 + (64 N)2
= 397 N
and this force is directed at an angle θ above the horizontal, where θ is given by
θ = tan
−1
Fseat, vert
Fseat, horiz
!
= tan−1
392 N
64 N
= 81◦
The force of the seat has magnitude 397 N and is directed at 81◦ above the horizontal.
128
CHAPTER 5. FORCES AND MOTION II
r
R
R
r=2R
Figure 5.22: Satellite orbits the Earth in Example 14.
5.2.3
Newton’s Law of Gravity (Optional for Calculus–Based)
14. A satellite of mass 300 kg is in a circular orbit around the Earth at an altitude
equal to the Earth’s mean radius (See Fig. 5.22.) Find (a) the satellite’s orbital
speed, (b) the period of its revolution, and (c) the gravitational force acting on
it. Use: REarth = 6.37 × 106 m and MEarth = 5.98 × 1024 kg. [Ser4 6-6]
(a) (Comment: This was the way the problem was originally stated. I will find the answers
in a different order!)
We are told that the height of satellite above the surface of the Earth is equal to the
Earth’s radius. This means that the radius of the satellite’s orbit is equal to twice the radius
of the Earth. Since the mean radius of the Earth is R = 6.37 × 106 m, then the orbit radius
is
r = 2R = 2(6.37 × 106 m) = 1.27 × 107 m .
The satellite is always at this distance from the center of the Earth; Newton’s law of gravitation tells us the force which the Earth exerts on the satellite:
Fgrav = G
=
msat MEarth
r2
6.67 × 10−11
= 7.42 × 102 N
N·m2
kg2
(300 kg)(5.98 × 1024 kg)
(1.27 × 107 m)2
This force is always directed toward the center of the Earth. Since this is the only force
which acts on the satellite, it is also the (net) centripetal force on it:
Fcent =
mv 2
= 7.42 × 102 N
r
We can now find the speed of the satellite. It is:
v2 =
rFcent
(1.27 × 107 m)(7.42 × 102 N)
=
= 3.14 × 107
m
(300 kg)
m2
s2
129
5.2. WORKED EXAMPLES
which gives
v = 5.60 × 103
So the satellite’s orbital speed is 5.60 × 103
m
s
.
m
.
s
(b) Recall that the speed of an object in uniform circular motion is related to the period
and radius by:
2πr
v=
T
From this we get the period of the satellite’s orbit:
2πr
v
2π(1.27 × 107 m)
=
(5.60 × 103 ms )
T =
= 1.42 × 104 s = 3.96 hr
The period of the satellite is 3.96 hr.
(c) The answer to this part has been found already! The gravitational force acting on the
satellite is 7.42 × 102 N.
130
CHAPTER 5. FORCES AND MOTION II